Question

With the following information given: y=x^2 , how do i find the coordinates of the vertex of this PARABOLA?

Answer 1

x^2 cannot be negative - minimum value = 0 at (0,0)
vertex is at (0,0)

Answer 2

how did you find out the vertex coordinates???

Answer 3

can you take step by step please.

Answer 4

\[y=4x^2\]

Answer 5

you can use that example ^^

Answer 6

the vertex will be at the minimum or maximum value of y
in this case minimum value is 0 as a square cannot have a real neagative value
when y = 0 x = 0
therefore the vertex is at (0,0)

Answer 7

hmm use y=4x^2 this time

Answer 8

y = 4x^2
same argument holds for this - minmumm value of 4x^2 = 0
and when y = 0 x= 0
vertex is at (0,0)

Answer 9

what do you mean minimum value of 4x^2 = 0 i don’t understand that!!

Answer 10

well a square value ie in this case x^2 cannot be negative eg -2 & -2 = +4
3 * 3 = 9
so the lowest value a square can be is 0
0 * 0 = 0

and anything times 0 eg 4 * 0 = 0

therefore lowestt value of y^2, 5x^2 ax^2 etc is 0

Answer 11

ok well what if it’s y=x^2 + 12

Answer 12

in this case the vertex is at y = 0 + 12 = 12
ie at (0,12)

for the general form
ax^2 + bx + c
thex coordinate of the vertex
is - b / 2a

Answer 13

i haven’t taken that equation yet
i only know y=a(x-h) +k

Answer 14

eg

for x^2 + 3x - 6
the vertex
is at x = -3/2*1 = -3/2
when x = -3/2 y = (-3/2)^2 + 3 (-3/2) - 6 = -9/4 - 9/2 - 6 = -51/4

so vertex is at (-3/2, -51/4)

Answer 15

but idk that equation

Answer 16

i was only taught for now the y = a(x-h)^2 +k

Answer 17

oh ok this the the vertex form of a quadratic
when x = h the a(x-h)^2 = 0 and then y = k

Answer 18

ok so whenever i see any quadratic equation i grab the H value and the K value and then i WHATEVER THE H AND K VALUES ARE, THATS THE VERTEX COORDINATES?

Answer 19

and in this form the vertex is (h,k)

Answer 20

yes

Answer 21

oh but its -h, do i ignore the negative sign on the h?

Answer 22

say i have y=5(x-3)^2 , the vertex coordinates are (3,0) OR (-3,0)

Answer 23

in this case the vertex is at
(3,0)

if y = 5(x+3)^2 the vertex is at (-3,0)

because the -3 makes (x+3) = 0

Answer 24

ohhhhhh i sees now bro.

Answer 25

so just to get this clear:

Answer 26

i’m gonna write out a conclusion and i want you to tell me if it’s correct. ok?

Answer 27

ok

Answer 28

To find out the coordinates of a vertex algebraically, you must take the value of h and replace it in for X, and take k and replace it in for Y. if h is positive, the number will be negative. if h is negative, the value will be positive. If there is no H, the vertex point x=0 . if there is no k, y=0 in the vertex.

Answer 29

example: y=3(x-3)^2 + 17
vertex point: (3,17) <<<<--- am i right??

Answer 30

absolutely right
and for
y=3(x+3)^2 - 17
what is the vertex?

Answer 31

(3,-17) ?

Answer 32

nearly : -17 is right but wahat about the x coordinate?

Answer 33

ohhhh yeah its -3 i forgot yup

Answer 34

cause h is naturally negative

Answer 35

so 2 negatives form a positive.

Answer 36

yup
or another way to look at it is (x+3) must be = 0
therefore x must be -3

Answer 37

ohhh thats a good way too.
anyways, thank you so much for your help. if i could give you 100 medals, i would. i really appreciate it, i now understand it because of you! thanks :D

Answer 38

thats ok

Answer 39

set first derivative in terms of x equal to zero and solve for x.
dx/d= 2x for 2x=0, x=0, plug in 0 for the original equation and y also equals zero,
answer: (0,0)

Answer 40

ty